Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int pre[200005];
int lowbit(int n) { return n & -n; }
void update(int n, int x) {
  for (; n < 2e5 + 5; n += lowbit(n)) pre[n] += x;
}
int presum(int n) {
  int ret = 0;
  for (; n; n -= lowbit(n)) ret += pre[n];
  return ret;
}
int l[200005];
int r[200005];
int main() {
  int sy, a, b, n, q;
  scanf("%d %d %d", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d %d", &l[i], &r[i]);
  l[n + 1] = r[n + 1] = 1e9 + 10;
  for (int i = 1; i <= n; i++) update(i, r[i] - l[i]);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    double a0 = a + (x - a) * ((double)(0 - sy) / (y - sy));
    double b0 = b + (x - b) * ((double)(0 - sy) / (y - sy));
    double sum = 0;
    int step = upper_bound(r + 1, r + n + 1, b0) - (r + 1) + 1;
    sum += presum(step - 1);
    if (l[step] <= b0) sum += b0 - l[step];
    step = upper_bound(r + 1, r + n + 1, a0) - (r + 1) + 1;
    sum -= presum(step - 1);
    if (l[step] <= a0) sum -= a0 - l[step];
    printf("%f\n", sum * (b - a) / (b0 - a0));
  }
  return 0;
}